Problem Description
Alice gets two sequences A and B. A easy problem comes. How many pair of sequence A' and sequence B' are same. For example, {1,2} and {1,2} are same. {1,2,4} and {1,4,2} are not same. A' is a subsequence of A. B' is a subsequence of B. The subsequnce can be not continuous. For example, {1,1,2} has 7 subsequences {1},{1},{2},{1,1},{1,2},{1,2},{1,1,2}. The answer can be very large. Output the answer mod 1000000007.
Input
The input contains multiple test cases.
For each test case, the first line cantains two integers N,M(1≤N,M≤1000). The next line contains N integers. The next line followed M integers. All integers are between 1 and 1000.
For each test case, the first line cantains two integers N,M(1≤N,M≤1000). The next line contains N integers. The next line followed M integers. All integers are between 1 and 1000.
Output
For each test case, output the answer mod 1000000007.
Sample Input
3 2
1 2 3
2 1
3 2
1 2 3
1 2
Sample Output
2
3
-----AC------
关于状态转移方程
if(a[i]==b[j])
dp[i][j]=(dp[i-1][j]+dp[i][j-1]+1);
else
dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]);
dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]);
例：
序号：1 2 3 4
a[i]:1 2 3 4
b[j]:1 2 3 1
a[i]:1 2 3 4
b[j]:1 2 3 1
假设（从dp[i][j-1]到dp[i][j]）：
if(a[i]==b[j]) 前面的j-1个数有dp[i][j-1]种可能，前面j-1个数和p组合又有dp[i][j-1]种可能，加上a[i]==b[j]，dp[i][j]=dp[i][j-1]*2+1；
else dp[i][j]=dp[i][j-1]；
推导得
a： 1 2 3 4
b： 1 2 3 1
dp[j]： 1 3 7 7
那么就会发现这样会忽略前i-1个数的子序列，所以为了不忽略，下面给出推导正确过程
首先，前面dp[i][j-1]，+1是正确的，错在（前面j-1个数和p组合又有dp[i][j-1]种可能,忽略前i-1个数的子序列），实际上可以把a，b倒过来（a[i]==b[j]）
例如：
a： 1 2 。。。
b: 1 。 。。。
看成
b: 1 2 。。。j
a: 1 .。。。。i
这样对和p组合可以看成dp[i-1][j]
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<cmath> 5 using namespace std; 6 #define MAXN 1005 7 #define INF 9999999999 8 #define mod 1000000007 9 int dp[MAXN][MAXN]; 10 int main() 11 { 12 int n,m; 13 while(cin>>n>>m) 14 { 15 int a[MAXN],b[MAXN],i,j; 16 memset(dp,0,sizeof(dp)); 17 for(i=1;i<=n;++i) 18 cin>>a[i]; 19 for(i=1;i<=m;++i) 20 cin>>b[i]; 21 for(i=1;i<=n;++i) 22 { 23 for(j=1;j<=m;++j) 24 { 25 if(a[i]==b[j]) 26 dp[i][j]=(dp[i-1][j]+dp[i][j-1]+1)%mod; 27 else 28 dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1])%mod; 29 if(dp[i][j]<0) 30 dp[i][j]+=mod; 31 } 32 } 33 // for(i=1;i<=n;++i) 34 // { 35 // for(j=1;j<=m;++j) 36 // cout<<dp[i][j]<<" "; 37 // cout<<endl; 38 // } 39 cout<<dp[n][m]<<endl; 40 } 41 return 0; 42 }